I am new to conditional probability and was wondering if anyone could help me out understanding this probability better.
$$P(aX+bZ\geq t \mid X)$$ Here, X and Z are i.i.d N(0,1)
To be more specific, I tried to calculate $Var(P(aX+bZ\geq t \mid X))$ with respect to X, but I can't make a move without truly understanding what the inner really means. Any help is greatly appreciated.
By the Doob-Dynkin lemma $\mathbb P\{aX+bZ\ge t\,|\,\,X\}$, or more generally, any conditional expectation w.r.t. $X$, in particular, $$\tag{1} \mathbb E[f(X,Z)\,|\,X] $$ is a deterministic function of the random variable $X\,.$ Here $f$ is a deterministic function of $X$ and $Z$. Let's write this as $$ \mathbb E[f(X,Z)\,|\,X]=g(X)\,. $$
When $X$ and $Z$ are independent then $g$ is simply $$ g(x)=\mathbb E[f(x,Z)]\,\quad x\in\mathbb R\,. $$ In other words, the conditional expectation (1) is essentially an unconditional expectation that depends on a real parameter $x\,.$ We get (1) by "plugging" $X$ in for $x$ after taking that unconditional expectation.
When we apply all this to your $\mathbb P\{aX+bZ\ge t\,|\,\,X\}$ it turns out that we only have to calculate the probability (for arbitrary $x\in\mathbb R$) $$\tag{2} \mathbb P\{ax+bZ\ge t\} $$ which is the complement of the CDF of the normal distribution with mean $ax$ and standard deviation $b^2\,.$